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# hw5_sol - Homework 5 Solution March 3 2009 Solution to...

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Homework 5 Solution March 3, 2009 Solution to Exercise 4, page 155 : (a) maximize z = - 5 x 1 + 2 x 2 subject to - x 1 + x 2 ≤ - 2 2 x 1 + 3 x 2 5 x 1 , x 2 0 . We transform the problem to a standard form, and obtain maximize z = - 5 x 1 + 2 x 2 subject to - x 1 + x 2 + x 3 = - 2 2 x 1 + 3 x 2 + x 4 = 5 x 1 , x 2 , x 3 , x 4 0 . Assigning dual variables y 1 and y 2 to the equality constraints, we obtain that the dual is given by minimize w = - 2 y 1 + 5 y 2 subject to - y 1 + 2 y 2 ≥ - 5 y 1 + 3 y 2 2 y 1 0 , y 2 0 . (b) minimize z = 6 x 1 + 3 x 2 subject to 6 x 1 - 3 x 2 + x 3 2 3 x 1 + 4 x 2 + x 3 5 x 1 , x 2 , x 3 0 . Again, we write the problem in the standard form minimize z = 6 x 1 + 3 x 2 subject to 6 x 1 - 3 x 2 + x 3 - x 4 = 2 3 x 1 + 4 x 2 + x 3 - x 5 = 5 x 1 , x 2 , x 3 , x 4 , x 5 0 . 1

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We next assign dual variables y 1 and y 2 to the equality constraints, we obtain that the dual is maximization and it is given by maximize w = 2 y 1 + 5 y 2 subject to 6 y 1 + 3 y 2 6 - 3 y 1 + 4 y 2 3 y 1 + y 2 0 - y 1 0 , - y 2 0 , or equivalently maximize w = 2 y 1 + 5 y 2 subject to 6 y 1 + 3 y 2 6 - 3 y 1 + 4 y 2 3 y 1 + y 2 0 y 1 0 , y 2 0 , (c) maximize z = x 1 + x 2 subject to 2 x 1 + x 2 = 5 3 x 1 - x 2 = 6 x 1 , x 2 unrestricted . To transform the problem to standard form, we need to handle x 2 since it is unre- stricted. We write x 2 = x 3 - x 4 , with x 3 , x 4 0. We obtain maximize z = x 1 + x 3 - x 4 subject to 2 x 1 + x 3 - x 4 = 5 3 x 1 - x 3 + x 4 = 6 x 1 , x 3 , x 4 0 . We assign dual variables y 1 and y 2 to the equality constraints, and we obtain the dual: minimize w = 5 y 1 + 6 y 2 subject to 2 y 1 + 3 y 2 1 y 1 - y 2 1 - y 1 + y 2 ≥ - 1 y 1 and y 2 unrestricted . The last inequality is equivalent to y 1 - y 2 1, which together with y 1 - y 2 1, 2
yields y 1 - y 2 = 1. Thus, the dual is given by minimize w = 5 y 1 + 6 y 2 subject to 2 y 1 + 3 y 2 1 y 1 - y 2 = 1 y 1 and y 2 unrestricted . Solution to Exercise 5, page 163 : The given LP: maximize z = 2 x 1 + 4 x 2 + 4 x 3 - 3 x 4 subject to x 1 + x 2 + x 3 = 4 x 1 + 4 x 2 + x 4 = 8 x 1 , x 2 , x 3 , x 4 0 is already in the standard form. We assign y 1 and y 2 to the equality constraints and obtain the dual: minimize w = 4 y 1 + 8 y 2 subject to y 1 + y 2 2 ( x 1 ) y 1 + 4 y 2 4 ( x 2 ) y 1 4 ( x 3 ) y 2 ≥ - 3 ( x 4 ) . The variables to the right indicate the correspondence of the primal variables and dual constraints. We are given that x 3 and x 4 are starting variables, and that the optimal tableau is: Basic x 1 x 2 x 3 x 4 Solution z 2 0 0 3 16 x 3 0 . 75 0 1 - 0 . 25 2 x 2 0 . 25 1 0 0 . 25 2 By method 1, we ﬁnd the dual optimal by looking at the reduced costs of x 3 and x 4 (starting variables), which are 0 and 3, and we look at the inequalities in the dual that are formed from the primal columns of

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hw5_sol - Homework 5 Solution March 3 2009 Solution to...

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